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Home/ Questions/Q 3335742
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T00:03:11+00:00 2026-05-18T00:03:11+00:00

I’d like to create a file with the name passenger_wsgi.py on a remote host.

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I’d like to create a file with the name passenger_wsgi.py on a remote host. I’d like to use the following string to create the file’s content:

'''
import sys, os

sys.path.insert(0, "/ruby/%s/www/%s/django-projects")
sys.path.insert(0, "/ruby/%s/www/%s/django-projects/project")

import django.core.handlers.wsgi
os.environ['DJANGO_SETTINGS_MODULE'] = 'project.settings'
application = django.core.handlers.wsgi.WSGIHandler()
''' % (user,host,user,host)

The user and host variables would be parameters of the fabric function.

I’m a total newbie to any sort of file manipulation in python, but also I’m not really sure what the procedure should be in fabric. Should I be creating the file locally and then uploading it with fabric’s put command (and removing the local version afterwards)? Should I be creating the file on the remote host with an appropriate bash command (using fabric’s run)? If so, then how is it best to deal with all the ” and ‘ in the string – will fabric escape it? Or should I be tackling this in some different manner?

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  1. Editorial Team
    Editorial Team
    2026-05-18T00:03:12+00:00Added an answer on May 18, 2026 at 12:03 am

    You could use append() or upload_template() functions from fabric.contrib.files

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